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Learn/ A Level Mathematics/ Past Papers/ 2020/ May/June 2020 /Paper 3 (Variant 1)

Pure Mathematics 3 — Variant 1

A Level Mathematics · 9709/31 · May/June 2020

Question Paper

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Mark Scheme

PDF · 369 KB · 13 pages

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Questions

7 of 10 questions shown here, checked against the paper. The others are in the PDF above — they contain diagrams or maths we could not yet verify, and we would rather show you the original than a transcription we are not sure of.

1 [4]
Find the set of values of xx for which 2(31−2x)<5x2(3^{1-2x}) < 5^{x}. Give your answer in a simplified exact form.
2
2(a) [4]
Expand (2−3x)−2(2 - 3x)^{-2} in ascending powers of xx, up to and including the term in x2x^{2}, simplifying the coefficients.
2(b) [1]
State the set of values of xx for which the expansion is valid.
3 [6]
Express the equation tan⁡(θ+60∘)=2+tan⁡(60∘−θ)\tan(\theta + 60^\circ) = 2 + \tan(60^\circ - \theta) as a quadratic equation in tan⁡θ\tan \theta, and hence solve the equation for 0∘⩽θ⩽180∘0^\circ \leqslant \theta \leqslant 180^\circ.
4
The curve with equation y=e2x(sin⁡x+3cos⁡x)y = \mathrm{e}^{2x}(\sin x + 3\cos x) has a stationary point in the interval 0⩽x⩽π0 \leqslant x \leqslant \pi.
4(a) [4]
Find the xx-coordinate of this point, giving your answer correct to 2 decimal places.
4(b) [2]
Determine whether the stationary point is a maximum or a minimum.
5
5(a) [3]
Find the quotient and remainder when 2x3−x2+6x+32x^{3} - x^{2} + 6x + 3 is divided by x2+3x^{2} + 3.
5(b) [5]
Using your answer to part (a), find the exact value of ∫132x3−x2+6x+3x2+3 dx\int_{1}^{3} \frac{2x^{3} - x^{2} + 6x + 3}{x^{2} + 3} \, \mathrm{d}x.
8
A certain curve is such that its gradient at a point (x,y)(x, y) is proportional to yxx\frac{y}{x\sqrt{x}}. The curve passes through the points with coordinates (1,1)(1, 1) and (4,e)(4, \mathrm{e}).
8(a) [8]
By setting up and solving a differential equation, find the equation of the curve, expressing yy in terms of xx.
8(b) [1]
Describe what happens to yy as xx tends to infinity.
9
With respect to the origin OO, the vertices of a triangle ABCABC have position vectors OA→=2i+5k\overrightarrow{OA} = 2\mathbf{i} + 5\mathbf{k}, OB→=3i+2j+3k\overrightarrow{OB} = 3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} and OC→=i+j+k\overrightarrow{OC} = \mathbf{i} + \mathbf{j} + \mathbf{k}.
9(a) [3]
Using a scalar product, show that angle ABCABC is a right angle.
9(b) [2]
Show that triangle ABCABC is isosceles.
9(c) [4]
Find the exact length of the perpendicular from OO to the line through BB and CC.

Published by Cambridge Assessment International Education. Reproduced for study use.